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Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{25}=5\)
\(\sqrt{5\sqrt{3+5\sqrt{48}-10\sqrt{7+4\sqrt{3}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48}-10\sqrt{4+2.2\sqrt{3+\left(\sqrt{3}\right)^2}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-10.\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-10.\sqrt{\left(2+\sqrt{3}\right)}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-20+10}\sqrt{3}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{28+10}\sqrt{3}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}\right)^2}+2.5.\sqrt{3}+5^2}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}+5\right)^2}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}+5\right)}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{3+10}}}\)
=\(10\sqrt{3+10}\)
=\(\sqrt{10\left(\sqrt{3+1}\right)}\)
Bài làm:
a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)
\(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\)
\(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\)
\(A=4+2\sqrt{3}+5\sqrt{3}-1\)
\(A=3+7\sqrt{3}\)
b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)
\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)
\(A=3+\sqrt{5}-\sqrt{5}+\sqrt{2}-\sqrt{2}-1\)
\(A=2\)
Phần b mình viết nhầm tên thành A, bn sửa thành B nhé
c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(C=\sqrt{3}-1-2-\sqrt{3}\)
\(C=-3\)
a: \(=6\sqrt{2}-12\sqrt{3}-10\sqrt{2}+12\sqrt{3}=-4\sqrt{2}\)
b: \(=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\sqrt{4-3}=1\)
Rút gọn:
\(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{2\sqrt{3+\sqrt{5-\left(1+2\sqrt{3}\right)}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{\sqrt{3+\sqrt{5-\left(1+2\sqrt{3}\right)}}.\left(\sqrt{6}-\sqrt{2}\right)}{2}\)
= \(\dfrac{\sqrt{\left[3+\sqrt{5-\left(1+2\sqrt{3}\right)}\right].6}-\sqrt{\left[3+\sqrt{5-\left(1+2\sqrt{3}\right)}\right].2}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{5-1-2\sqrt{3}}\right).6}-\sqrt{\left(3+\sqrt{5-1-2\sqrt{3}}\right).2}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{4-2\sqrt{3}}\right).6}-\sqrt{\left(3+\sqrt{4-2\sqrt{3}}\right).2}}{2}\)
= \(\dfrac{\sqrt{\left[3+\sqrt{\left(1-\sqrt{3}\right)^2}\right].6}-\sqrt{\left[3+\sqrt{\left(1-\sqrt{3}\right)^2}\right].2}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{3}-1\right).6}-\sqrt{\left(3+\sqrt{3}-1\right).2}}{2}\)
= \(\dfrac{\sqrt{\left(2+\sqrt{3}\right).6}-\sqrt{\left(2+\sqrt{3}\right).2}}{2}\)
= \(\dfrac{\sqrt{12+6\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}}{2}\)
= \(\dfrac{3+\sqrt{3}-\left(1+\sqrt{3}\right)}{2}\)
= \(\dfrac{3+\sqrt{3}-1-\sqrt{3}}{2}\)
= \(\dfrac{2}{2}\)
= \(1\)
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+\sqrt{48}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2-\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20+10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
= 5
\(\dfrac{\sqrt{3}-\sqrt{5+\sqrt{24}}+\sqrt{\sqrt{72}+11}}{\sqrt{6+\sqrt{20}}+\sqrt{2}-\sqrt{7+\sqrt{40}}}\)
\(=\dfrac{\sqrt{3}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}-\sqrt{3}+3+\sqrt{2}}{\sqrt{5}+1+\sqrt{2}-\sqrt{2}-\sqrt{5}}\)
\(=3\)